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In a Young's double-slit experiment, light of wavelength 500 nm is used. The slit separation is 1 mm and the screen is 2 m away. A thin slab of thickness t = 3 micrometers and refractive index mu = 3/2 is placed in front of one slit. What is the intensity at the centre of the screen O? (Intensity from each slit alone = I0)

1. 3 * I0
2. 4 * I0
3. I0 / 2
4. 2 * I0

Correct answer: 4 * I0

Solution

The slab introduces extra optical path = (mu-1)*t = (3/2-1)*3 micrometers = 0.5*3 = 1.5 micrometers. Phase difference at centre O = 2*pi*(mu-1)*t / lambda = 2*pi*1.5/0.5 = 6*pi. Since 6*pi = 3*(2*pi), the phase difference is an integer multiple of 2*pi, giving constructive interference. Maximum intensity I_max = (sqrt(I0)+sqrt(I0))² = 4*I0.

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